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Kindly refer to the figure, AB is the tangent to the circle with cente

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Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  23 May 2020, 19:42
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Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees. If OA =4 cm, and OB = 3cm. What is the radius of the circle?
A. 1.6
B. 2.1
C. 2.4
D. 3.2
E. 3.5

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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  23 May 2020, 19:47
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1/2×3×4=1/2×5×r
r =2.4

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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  23 May 2020, 19:51
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answer C
Let D be point radius from O touches tangent
triangle OAD and triangle OAB are simmilar
hence corresponding ratio of sides equal
radius/4=3/5
radius=2.4
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  23 May 2020, 19:54
If we join the center of the circle with the point of tangency, we get a right angle. Let the point of tangency be T and the line be OT. OT is the radius of the circle.
We can see the AB=5 (3-4-5 pythagorean triplet)
Area of triangle AOB= 1/2 * AO* AB= 1/2* 3*4= 6
Also, Area= 1/2 * OT* AB
so 6= 1/2 * OT* 5
OT= 12/5= 2.4.
Answer: C
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Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  Updated on: 25 May 2020, 09:01
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given OB is 3 cm so radius of circle has to be < 3cm D &E are out
now for
∆ AOB ; side AB = 5 AO = 4 and BO = 3
∆ AOD OD = r and AO = 4
OD/OA= OB/AB
=> r/4 = 3/5
=> r= 12/5 ; 2.4
OPTION C



Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees. If OA =4 cm, and OB = 3cm. What is the radius of the circle?
A. 1.6
B. 2.1
C. 2.4
D. 3.2
E. 3.5

Originally posted by Archit3110 on 23 May 2020, 19:57.
Last edited by Archit3110 on 25 May 2020, 09:01, edited 2 times in total.
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  23 May 2020, 20:18
Given: Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees.
Asked: If OA =4 cm, and OB = 3cm. What is the radius of the circle?

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Since \(\triangle\) AOB & \(\triangle\) AOD are similar; Angle A & \(90^0\) are common.
\(\frac{OD}{AO} = \frac{r}{4} = \frac{OB}{AB} = \frac{3}{5}\)
Radius of circle = \(r = \frac{12}{5} = 2.4\)

IMO C
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  23 May 2020, 20:39
IMO C

AB is the tangent to the circle with center O,
OA =4 cm, and OB = 3cm.

Lets Draw radius OD from O at point of contact of tangent to circle.(Shown in figure)
Tangent is perpendicular to radius at point of contact.
So ∠ ADO= 90

In triangle AOB & ADO
∠AOB=∠ADO=90
∠OAB=∠DAO (Common)
Both triangles are similar
OB/AB= OD/AO
OD= OB x AO/AB = 3 X 4 /5 = 2.4

C. 2.4
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  23 May 2020, 23:34
In triangle AOB, <AOB = 90 deg
OA = 4, OB = 3
using rule of AB^2 = OA^2 + OB^2
AB = 5

Now radius of circle is perpendicular to tangent
Hence considering rule of triangle area

1/2 X OA X OB = 1/2 X R X AB
R = (OA X OB) / AB = 3X4/5 = 2.4
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  24 May 2020, 02:12
Quote:
Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees. If OA =4 cm, and OB = 3cm. What is the radius of the circle?
A. 1.6
B. 2.1
C. 2.4
D. 3.2
E. 3.5


ab^2 = 4^2+3^2
ab=5
area ABO(considering base BO) = 1/2*3*4=6
area ABO(considering base AB) = 1/2*5*h=6 (h=point where circle and ab intersect/meet)
therefore h=12/5
and h=r
so r= 2.4(ans)
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  24 May 2020, 03:14
Area of triangle AOB = \(\frac{1}{2} * base * height \)
=\(\frac{1}{2}\) * 3 * 4 = 6 sq. cm

We know , radius of circle which joins centre to tangent makes right angle with circumference.

AB = \(\sqrt{AO^2 + OB^2}\) = \(\sqrt{25}\) = 5 cm (by pythagoras theorem)

Also, Area of triangle AOB = \(\frac{1}{2}\) * AB * radius of circle
=> 6 = 2.5 * radius of circle
=> Radius of circle = \(\frac{6}{2.5}\) = 2.4 cm

Hence, OA is C.
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  24 May 2020, 03:42
Answer: OPTION C - 2.4

As AB is a tangent and radius of a circle is always perpendicular on a tangent, hence we can draw a perpendicular from O to AB, as OD.

Using Pythagoras theorem,

\(AB ^ 2 = (OA) ^ 2 + (OB) ^ 2\) = (4 ^ 2) + (3 ^ 2)
=> AB = 5

\(Area of traingle = 1/2 * (base * height)\)

As triangle OAB is a right triangle and has two perpendiculars so,

=> \(1/2 * (OA * OB) = 1/2 * (AB * OD)\)
=> \(1/2 * 4 * 3 = 1/2 * 5 * OD\)
=> \(OD = 12 / 5\) = 2.4
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  24 May 2020, 04:37
Let's say that the points where the tangent is touching the circle be x.
Angle OXB = 90, and angle XBO is 45. Sin 45 = perpendicular/hypotenuse.
Hypotenuse is 3, so 1/√2 = ox/3, OX = 3/√2
Which is 2.08 so 2.1

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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  24 May 2020, 10:49
Answer : Option C ( 2.4cm )

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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  24 May 2020, 13:03
C .

Let point of intersection of circle with Tangent AB be ' D'.
Also, AB = 5 for Right Angle Triangle AOB with legs OA = 4 & OB = 3

Area of Triangle AOB = 1/2 x OB x OA = 1/2 x AB x OD ( For Base AB , OD is the altitude )
Radius = OD = 12 /5 = 2.4 cm
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink] New post  10 Mar 2023, 22:33
Equate the areas.
1/2*4*3 =6
Consider 5 as the base and draw a radius perpendicular to the base 5
Equate the areas.
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Re: Kindly refer to the figure, AB is the tangent to the circle with cente [#permalink]
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